Problem: $g(x)=\begin{cases} 2^x-1&\text{for }-8\leq x<1 \\\\ \sqrt x&\text{for }x\geq1 \end{cases}$ Find $\lim_{x\to 4}g(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $2$ (Choice C) C $15$ (Choice D) D The limit doesn't exist.
Answer: Let's find the limit as $x$ approaches $4$. We will use the fact that $g(x)=\sqrt x$ for $x$ -values larger than $1$. $\begin{aligned} &\phantom{=}\lim_{x\to 4}g(x) \\\\ &=\lim_{x\to 4}\sqrt x \\\\ &=\sqrt4&\gray{\text{Direct substitution}} \\\\ &=2 \end{aligned}$ In conclusion, we found that $\lim_{x\to 4}g(x)=2$.